Question

Let $y\left(x\right)$ be a solution of the differential equation $(1+e^x)y^{\prime }+y{e}^x=1$. If $y\left(0\right)=2$, then which of the following statements is (are) true?

• A. $y(-4)=0$
• B. $y(-2)=0$
• C. $y\left(x\right)$ has a critical point in the interval $(-1,0)$
• D. $y\left(x\right)$ has no critical point in the interval $(-1,0)$

Answer 1

Answer: (A), (C)

$y(-4)=0$
$y\left(x\right)$ has a critical point in the interval $(-1,0)$

$(1+e^x)\frac{dy}{dx}+y{e}^x=1$

$\Rightarrow \frac{dy}{dx}+\frac{e^x}{1+e^x}y=\frac{1}{1+e^x}$

$I.F=e^{\int \frac{e^x}{1+e^x}dx}=e^{\ln (1+e^x)}=1+e^x$

Thus solution is,

$y(1+e^x)=\int dx+c=x+c$

Given $y\left(0\right)=2\Rightarrow c=4$

$\Rightarrow y=\frac{x+4}{1+e^x}$

Clearly $y(-4)=0$

Also $\frac{dy}{dx}=\frac{(1+e^x)-e^x(x+4)}{(1+e^x{)}^2}$

For critical point $\frac{dy}{dx}=0\Rightarrow (1+e^x)-e^x(x+4)=0$

Let $g\left(x\right)=(1+e^x)-e^x(x+4)$

$\therefore g(-1)=1-\frac{2}{e}>0$ and $g\left(0\right)=2-4=-2<0$

Hence $y\left(x\right)$ posses a critical point in the interval $(-1,0)$